24. 两两交换链表中的节点

给你一个链表，两两交换其中相邻的节点，并返回交换后链表的头节点。你必须在不修改节点内部的值的情况下完成本题（即，只能进行节点交换）。

方法一、递归

首先定义递归终止条件：

• head.next不存，代表链表结束了
• head.next.next不存在，表示不能两两配对

Swift

func swapPairs(_ head: ListNode?) -> ListNode? {
        //递归实现，抽象
        //终止条件就是当结束or只剩一个时，终止
        if head == nil || head?.next == nil {
            return head
        }
        
        let newHead:ListNode? = head?.next
        head?.next = swapPairs(newHead?.next)
        newHead?.next = head
        
        return newHead
    }

OC

- (ListNodeOC * _Nullable)swapPairs:(ListNodeOC * _Nullable)head {
    //递归终止条件
    if (!head.next || !head.next.next) {
        return head
    }
    
    ListNodeOC *freshHead = [ListNodeOC alloc] initWithVal:0];
    while(head.next && head.next.next) {
        head.next = swapPairs(freshHead.next);
        freshHead.next = head;
    }
    
    return freshHead;
}

方法二、迭代

用到了解决链表问题的三把斧：哑巴节点、栈、前后指针，用tempNode标记，按步进为2向后迭代，依次两两交换

Swift

func swapPairs(_ head: ListNode?) -> ListNode? {
        //哑巴节点
        let dummyNode = ListNode(0)
        dummyNode.next = head;
        
        var tempNode:ListNode? = dummyNode
        
        while tempNode?.next != nil && tempNode?.next?.next != nil {
            let fir = tempNode?.next
            let sec = tempNode?.next?.next
            //1->3
            fir?.next = sec?.next
            //2->1
            sec?.next = fir
            //temp->next -> sec
            tempNode?.next = sec
            
            //update tempNode pointer
            tempNode = fir
        }
        
        return dummyNode.next
    }

OC

- (ListNodeOC * _Nullable)swapPairs:(ListNodeOC * _Nullable)head {
    //build dummy node
    ListNodeOC *dummyNode = [[ListNodeOC alloc] initWithVal:0];
    dummyNode.next = head;
    
    ListNodeOC *tempNode = dummyNode;
    while(tempNode.next && tempNode.next.next) {
        ListNodeOC *fir = tempNode.next;
        ListNodeOC *sec = tempNode.next.next;
        
        tempNode.next = sec;
        fir.next = sec.next;
        sec.next = fir;
        
        tempNoe = fir;
    }
    
    return dummyNode.next;
}